# Homework Help: 2-D vector analysis help

1. Feb 21, 2014

### princeton_wu

1. The problem statement, all variables and given/known data

http://imgur.com/wNusHOw

2. Relevant equations

I have the solutions and how they did it. THey took the deriv for the velocity vector, and then using t=0 and t=14, they found e=3.5 and f=-0.125

3. The attempt at a solution

I understand the math, but I don't understand why this is correct
1) why did they take the velocity vector?
2) if I plug e and f back into the position vector, and using t=14 i should get θ=0, ie, j-component is 0. But I don't. So what am I doing wrong?

2. Feb 21, 2014

### tiny-tim

welcome to pf!

hi princeton_wu! welcome to pf!
because the graph gives you the direction of the velocity vector
yes you do …

j-component = e + 2ft = 3.5 - 3.5 = 0

3. Feb 21, 2014

### princeton_wu

i mean the r vector; shouldn't the J-component of the r-vector be 0 @ t=14? this way, the angle @ t=14 would be 0.

4. Feb 21, 2014

### tiny-tim

i'm not following your reasoning

the graph shows that vj = 0 (because θ = 0) at t = 14, it says nothing about r

5. Feb 21, 2014

### princeton_wu

sorry, I'm confused too :tongue:

If you use t=14 and plug it in the original position vector, shouldn't the J-component of the position vector be 0? Reasoning being that for θ to be 0, the J-component has to be 0?

thanks for your patience tiny tim!

6. Feb 22, 2014

### tiny-tim

(just got up :zzz:)
but θ (given in the graph) is stated to be the angle of the velocity vector …

i don't understand what you think that has to do with the position vector

7. Feb 22, 2014

### princeton_wu

I mulled over it last night and I finallly got it. My problem stemmed from the fact that I didn't realize that a t vs θ graph is a velocity graph). Thanks Tim!